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JEE Main & Advanced Mathematics Differentiation Question Bank Derivative at a point Standard Differentiation

  • question_answer
    If \[y={{\tan }^{-1}}\left( \frac{{{x}^{1/3}}+{{a}^{1/3}}}{1-{{x}^{1/3}}{{a}^{1/3}}} \right)\], then \[\frac{dy}{dx}=\]      [DSSE 1986]

    A)          \[\frac{1}{3{{x}^{2/3}}(1+{{x}^{2/3}})}\]

    B)            \[\frac{1}{3{{x}^{2/3}}(1+{{x}^{2/3}})}\]

    C)            \[-\frac{1}{3{{x}^{2/3}}(1+{{x}^{2/3}})}\]

    D)            \[-\frac{a}{3{{x}^{2/3}}(1+{{x}^{2/3}})}\]

    Correct Answer: A

    Solution :

               \[y={{\tan }^{-1}}\left( \frac{{{x}^{1/3}}+{{a}^{1/3}}}{1-{{x}^{1/3}}.{{a}^{1/3}}} \right)={{\tan }^{-1}}({{x}^{1/3}})+{{\tan }^{-1}}{{a}^{1/3}}\]                    Þ \[\frac{dy}{dx}=\frac{1}{3{{x}^{2/3}}(1+{{x}^{2/3}})}\].


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